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$V = \mathbb{K}^n$, where $\mathbb{K}$ -- field. $V_1 = \{(x_1,\cdots,x_n)\in V\mid \sum_{i=1}^{n} a_ix_i = 1; a_1,\cdots a_n \in \mathbb{K}\}$. So, I should check that $V_1$ -- subspace.

At this point, I remembered the structure of dual space. So if I have the vector space $V$ and the field $\mathbb{K}$ (that a $1$-dim vec.space under the itself too), and I can construct $Hom(V,\mathbb{K})$. This is obviously a vector space, where the sum is determined for functions pointwise (by the sum on images of points). And it is clear that $Hom(V,\mathbb{K}) = \mathbb{K}^n$, where $n = |V|$ -- power of $V$-set. So we can see, that $Hom(V,\mathbb{K}) = V$. And if I find the subspace in $V_{1}^{*} \subset Hom(V,\mathbb{K})$, which will satisfy the same condition as $V_1$ I will proof that $V_1$ -- subspace too.

We can check $(f_1 + f_2)(x) = f_1(x) + f_2(x)$ and $(af)(x) = f(ax)$, for vectors $x \in V_1$. If $f_1(x) = \sum_{i=1}^{n} a_ix_i = 1 = f_2(x)$ $\Rightarrow$ $(f_1+f_2)(x) = 2$ -- contradiction.

Is this proof correct?

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  • $\begingroup$ @ThomasShelby the $V_1^{*}$ will not be a subspace under this linear condition $\endgroup$ – Just do it Dec 9 '18 at 4:27
  • $\begingroup$ You haven't defined $V_1^{*} $. $\endgroup$ – Thomas Shelby Dec 9 '18 at 5:04
  • $\begingroup$ @ThomasShelby The subset of $Hom(V,\mathbb{K})$, which satisfy the same linear condition, like $V_1$. This is should be ok, because $Hom(V,\mathbb{K})$ -- isomorphic vec.space to space $V$ $\endgroup$ – Just do it Dec 9 '18 at 5:07
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The path of least resistance is to consider the following hint.

Hint: Any subspace must contain the zero vector.

Does $V_1$ contain the zero vector?

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  • $\begingroup$ Yes, it is also possible proof, for some reason I first decided to go along a long path, because I was afraid of the arbitrariness of the field choice. But now I think that even if the field is finite, then there is no zero vector in the subset. $\endgroup$ – Just do it Dec 9 '18 at 4:54
  • $\begingroup$ But my proof is also correct ? $\endgroup$ – Just do it Dec 9 '18 at 4:55
  • $\begingroup$ @Arsenii I'm sorry. I can't grasp your proof completely :( $\endgroup$ – Thomas Shelby Dec 9 '18 at 5:48

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